Basel's problem: $5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)$ Golden ratio:$\phi$
$$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}=\zeta(2)\tag1$$
We'll known Basel's problem
$$\sum_{k=1}^{\infty}{1\over k^2}=\zeta(2)\tag2$$
How do I transform $(1)$ to $(2)?$
Extra information:
$$\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}=1\tag3$$
Nothing come to my mind! Any hints?
 A: You want to show that
$5\sum_{k=1}^{\infty}{\phi^k-1\over \phi^{2k}}\cdot{1\over k^2}
=\zeta(2)\tag1
$.
The function you need
is the dilogarithm,
defined by
$L_2(z)
=\sum_{k=1}^{\infty}\dfrac{z^k}{k^2}
$.
There is much useful info here:
https://en.wikipedia.org/wiki/Spence's_function
So you want to show that
$5(L_2(\phi^{-1})-L_2(\phi^{-2}))
=\zeta(2)
$.
Since
$\phi
=\dfrac{1+\sqrt{5}}{2}
$,
$\phi-1
=\dfrac{\sqrt{5}-1}{2}
$,
$1-\phi
=\dfrac{1-\sqrt{5}}{2}
=\dfrac{-1}{\phi}
$
and
$\dfrac{1}{\phi^2}
=\dfrac{(1-\sqrt{5})^2}{4}
=\dfrac{1-2\sqrt{5}+5}{4}
=\dfrac{3-\sqrt{5}}{2}
$,
this is equivalent to
$5(L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2}))
=\zeta(2)
$.
According to the Wikipedia page above,
$L_2(\dfrac{\sqrt{5}-1}{2})
=\dfrac{\pi^2}{10}-\ln^2(\dfrac{\sqrt{5}-1}{2})
$
and
$L_2(\dfrac{3-\sqrt{5}}{2}))
=\dfrac{\pi^2}{15}-\ln^2(\dfrac{\sqrt{5}-1}{2})
$.
Subtracting these,
the $\ln^2$ terms cancel out
and we get
$\begin{array}\\
L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2}))
&=\dfrac{\pi^2}{10}-\dfrac{\pi^2}{15}\\
&=\dfrac{\pi^2}{30}\\
\text{so}\\
5(L_2(\dfrac{\sqrt{5}-1}{2})-L_2(\dfrac{3-\sqrt{5}}{2})))
&=\dfrac{\pi^2}{6}\\
&=\zeta(2)\\
\end{array}
$
